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Find the sum 1 xx 2 xx .^(n)C_(1) + 2 xx 3xx .^(n)C_(2) + "….." + 2 xx (n+1) xx .^(n)C_(n). |
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Answer» Solution :METHODI : `1 xx 2 xx .^(n)C_(1)+ 2 xx 3 xx .^(n)C_(2) + "...." + n xx (n+1) xx .^(n)C_(n)` `= underset(r=1)OVERSET(n)sumr(r+1)..^(n)C_(r)` `= underset(r=1)overset(n)sum(r+1)[n..^(n-1)C_(r-1)]` `= n underset(r=1)overset(n)sum((r-1)+2).^(n-1)C_(r-1)` `= nxxunderset(r=1)overset(n)sum[(r-1)..^(n-1)C_(r-1)+2..^(n-1)C_(r-1)]` `= nxx n(n-1)underset(r=2)overset(n)sum.^(n-2)C_(r-2)+(2n)underset(r=1)overset(n)sum.^(n-1)C_(r-1)` `=n xx (n-1) xx 2^(n-2) + 2nxx 2^(n-1)` ` = n(n+3)xx2^(n-2)` Method II : We have `(1+x)^(n) = .^(n)C_(0) + .^(n)C_(1)x + .^(n)C_(2)x^(2) + "....." + .^(n)C_(n)x^(n)` Differentiatingw.r.t.x we get `n(1+x)^(n-1) = .^(n)C_(1) + 2 xx .^(n)C_(2)x + 3 xx .^(n)C_(3)x^(2)"....." + nxx .^(n)C_(n) xx x^(n-1)` Multipyingwith `x^(2)`,we have `n(1+x)^(n-1) x^(2) = .^(n)C_(1) x^(2) + 2 xx .^(n)C_(2)x^(3) + 3xx .^(n)C_(3)x^(4)"...."+n xx .^(n)C_(n)x^(n+1)` Differentiating w.r.t. ,we get `m(n-1)(1+x)^(n-2)x^(2) +2x xx n(1+x)^(n-1)` `= 2.^(n)C_(1)x + 2 xx 3xx .^(n)C_(2)x^(2) + 3 xx 4 xx .^(n)C_(3)x^(3) + "...." + n(n+1) xx .^(n)C_(n)x^(n)` Now putting ` x = 1`, we get `1 xx 2 xx .^(n)C_(1) + 2 xx 3 xx .^(n)C_(2) + "...." + n xx (n+1) xx .^(n)C_(n)` `= n (n+3) xx 2^(n-2)` |
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